Trace of matrix $A^{k}$ is

3^{k} + 1 + {(- 1)}^{k}

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2^{k} + 3^{k} - 2

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3^{k} - 2^{k} + 2

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2^{k} + 1

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Solution

The correct option is A $3^{k} + 1 + {(- 1)}^{k}$
$A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 2 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 2 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} (1 \times 1 + 2 \times 2 + 0 \times 0) & (1 \times 2 + 2 \times 1 + 0 \times 0) & 0 (1 \times 2 + 2 \times 1 + 0 \times 0) & (2 \times 2 + 1 \times 1 + 0 \times 0) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A^{3} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} (1^{2} + 2^{2}) & (1 \times 2 + 2 \times 1) & 0 (1 \times 2 + 2 \times 1) & (1^{2} + 2^{2}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 2 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 13 & 14 & 0 14 & 13 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ T r a c e (A^{2}) = 11, T r a c e (A^{3}) = 27 T r = 27; T r a c e (A^{4}) = 83$
$3^{k} + 1 + {(- 1)}^{k}$ gives
$k = 2 \Rightarrow 3^{2} + 1 + 1 = 11 k = 3 \Rightarrow 3^{3} + 1 - 1 = 27 k = 4 \Rightarrow 3^{4} + 1 + 1 = 83$

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